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So octonion set provides the largest normed division algebra, and starting with sedenions, Cayley-Dickson construction provides algebras with zero divisors.

From what I understand, it means there are pairs of non-null sedenion $(s_a,s_b)$ for which $ s_a·s_b = s_0$ is correct – where $s_0$ stands for the null sedenion $(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)$. But maybe I already misunderstood something here?

But given that higher order algebra gradually lose algebraic properties as the scale of dimension is climbed, are they any pair of non-null sedenion $(s_c, s_d)$ for which $s_c/s_0=s_d$ holds? That is, is it valid, at least in some cases, to divide by the null sedenion?

psychoslave
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    As long as "$a/b=c$" is shorthand for "$a=c\cdot b$" (order mattering here), the answer is clearly negative: just prove directly that $s_0\cdot t=t\cdot s_0=s_0$ for any sedenion $t$. (And if that's not what you mean by "$a/b=c$," what do you mean?) – Noah Schweber Mar 14 '22 at 16:16
  • On a side note, wheels define a type of algebra where zero division is somehow possible, although the definition of division and behavior or its operators in general is somehow changed. – psychoslave Apr 15 '22 at 16:30

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No it is not. In the Sedenion algebra you have not only one couple but many such that their product is 0, and 0 is still absorbant for the original product. So you cannot "divide" by 0 because there is no exclusivity.

ogerard
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